\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{2}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow A< 1+1-\frac{1}{2014}\)
\(\Rightarrow A< 2-\frac{1}{2014}< 2\)
Vậy A < 2 (đpcm)